Question

An integer is chosen at random from first 200 positive integers. Find the probability that the integer is divisible by 6 or 8.

Answer 1

Let S be the sample space. Then n(S) = 200
∴ Total number of elementary events = 200
Let A be the event in which the number selected is divisible by 6 and B be the event in which the number selected is divisible by 8.
Then A = {6, 12, 18, 24, ...198 },
B = { 8, 16, 24, 32, ...200}
and (A ∩ B) = {24, 48, 72, ...192}

Now, we have : \[n\left( A \right) = \frac{198}{6} = 33\]

\[n\left( B \right) = \frac{200}{8} = 25\]

\[n\left( A \cap B \right) = \frac{192}{24} = 8\]       [∵ LCM of 6 and 8 is 24]

\[\therefore P\left( A \right) = \frac{33}{200}, P\left( B \right) = \frac{25}{200} \text{ and } P\left( A \cap B \right) = \frac{8}{200}\]

Now, required probability = P(a number is divisible by 6 or 8)
                                       = P (A ∪ B)
                                       = P(A) + P(B) - P(A ∩ B)
                                       = \[\frac{33}{200} + \frac{25}{200} - \frac{8}{200} = \frac{33 + 25 - 8}{200} = \frac{50}{200} = \frac{1}{4}\]